The object will hit the ground after 5 seconds. You can rewrite the quadratic function as a quadratic equation set equal to zero to find the zeros of the function 0 = –16t2 + 80t + 0. You can factor or use the quadratic formula to get t = 0 and t = 5. Therefore, it is on the ground at t = 0 (time of launch) and then hits the ground at t = 5 seconds

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Step-by-step explanation:

289 espero que te resulte itil

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).